Maths - Formulae and Equations

Using Formula
1. A taxi firm charges £0.50 per miles plus a fixed rate of £2.00
  1. It costs £2 + £0.50 to travel 1 mile, It costs £2 + 2 x £0.50 to travel 2 miles ...
    1. So travelling 'n' miles will cost £2 + n x £0.50
      1. The formula is COST = £2 + (n x £0.50)
        Substitution
        What is the cost of hiring the taxi for 16 miles?
        C = £2 + (16 x £0.50)
        C = £2 + £8
        C = £10
2. A rectangle has a width of x and a length of 2x
  1. Perimeter = x + x + 2x + 2x
    1. P = 6x
Re-arranging Symbols
1. Collecting like terms
  1. To simplify an expression, we collect like terms
    1. 4x + 5x - 2 - 2x + 7
      1. The x terms can be collected together and the numbers can be collected together
        So 4x + 5x - 2x = 7x and 7 - 2 = 5
        This simplifies to 7x + 5
    2. x + 5 + 3x - 7 + 9x + 3 - 4x
      1. So x + 3x + 9x -4x = 9x and 5 + 3 - 7 =1
        This simplifies to 9x + 1
2. Different Terms
  1. You may have to simplify an equation with many different terms and letters
    1. 5a + 3b - 3a - 5c + 4b
      1. Collect like terms
        5a - 3a +3b +4b - 5c
        Simplify all terms
        2a + 7b - 5c
3. Multiplying out brackets
  1. 3(4x - 7)
    1. First Multiply: 3 x 4x = 12x
      1. Then multiply: 3 x -7 = -21
        Therefore: 3(4x - 7) = 12x - 21
  2. Remember whether the numbers are negative or positive
  3. Multiplying out two brackets
    1. (x+4)(x+3)
      1. FOIL
        x2 + 7x + 12
    2. Brackets and Powers
      1. (a-5) squared
        (a-5)(a-5)
        a2 - 10a + 25
Changing the subject of a formula
1. Arrange the formula C=2pi r to make r the subject
  1. Divide both sides by 2pi
    1. r = C/ 2pi
2. Rearrange the formula V = 4/3 pi r (3)
  1. Multiply by 3
    1. 3V = 4pi r (3)
      1. Divide by 4 pi
        3V/ 4 pi = r(3)
        Take the cube root of both sides
        r = cube root of 3V / 4pi
Simultaneous Equations
1. 2x + y = 7
  1. 3x - y = 8
    1. Add the equations
      1. 5x = 15 so...
        x = 3
        Substitute in...
        (2 x 3) + y = 7
        y = 7-6
        x = 3, y = 1
2. If the item you want to remove is two positives or two negatives you subtract, if they are one positive and one negative you add
Equations with Fractions
1. x/2 - 4 = 3
  1. +4 so x/2 = 7
    1. x2 so x = 14
Trial and Improvement
1. Always see to how many d.p the answer needs to be
  1. Remember to get as close to the final answer
Index Notation
1. Powers
  1. a squared = a x a
    1. b cubed = b x b x b
      1. 4d squared = 4 x d x d
2. Index Laws
  1. When multiplying you add the indices, when subtracting you divide the indices
    1. p3 x p7 = p10
      1. 4s3 x 3s2 = 12s5
Solving Equations
1. Using inverses
  1. x - 6 = 9
    1. x = 9+6
      1. x = 15
2. Unknowns on both side
  1. 3b + 4 = b + 12
    1. -b so 2b + 4 = 12
      1. -4 so 2b = 8
        /2 so b = 4
3. Equations with brackets
  1. 3(b+2) = 15
    1. 3 x b + 3 x 2 = 15
      1. 3b +6 = 15
        3b = 9
        b = 3

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	Criado por pondcott quase 11 anos atrás